example on dual space and annihilator space

Question

Say we have the set $W = \{(x_1 , \cdots , x_n) \in \mathbb{R}^n | x_1 + 2x_2 + \cdots + nx_n = 0\}$

I want to try finding the dual space and the annihilator space for the subspace $W$ of $\mathbb{R}^n$

I am not quite sure how to build from the definitions though.

I managed to find a basis for $W$ being :$\{2e_1 - e_2 , 3e_2 - 2e_3 , \cdots , ne_{n-1} - (n-1)e_n\}$

Thank you for the help!

Answer 1

To find the dual space, it suffices to determine its basis.

Definition (dual basis). The dual space of $V$, denoted $V'$, is the vector space of all linear functionals on $V$, i.e. linear maps from $V$ to $\mathbf{R}$. Let $v_1,\ldots, v_n$ is a basis of $V$, then the dual basis of $v_1,\ldots,v_n$ is the list $\varphi_1,\ldots, \varphi_n$ of elements of $V'$, where each $\varphi_j$ is the linear functional on $V$ such that $$\varphi_j(v_k)=\begin{cases} 1 & \text{if }k=j, \\ 0 & \text{if } k \ne j\end{cases}$$

And we have following result:

Theorem. For finite-dimensional vector space $V$ then the dual basis of a basis of $V$ is a basis of $V'$.

With this, you can determine the dual space $W'$ of $W$ or the dual space $(\mathbf{R}^n)'$ of $\mathbf{R}^n$.

For $W$ as a subspace of $\mathbf{R}^n$, the annihilator space $W^{0}$ of $W$ in $\mathbf{R}^n$ is the set $\{ \varphi \in (\mathbf{R}^n)': \varphi(w)=0 \; \forall w\in W\}$. One can show that $W^{0}$ is a subspace of $(\mathbf{R}^n)'$ so to find $W^{0}$, it suffices to find a basis of $W^{0}$.

From previous part, we know a basis $\varphi_1, \ldots, \varphi_n$ of $(\mathbf{R}^n)'$. You then need to find $a_1, \ldots, a_n \in \mathbf{R}$ such that $a_1\varphi_1+\ldots+a_n\varphi_n \in W^0$. Note that $\text{dim }W+\text{dim }W^0=\text{dim }\mathbf{R}^n=n$ and since $\text{dim }W=n-1$ so you would expect $W^{0}$ to be a subspace of $(\mathbf{R}^n)'$ with dimension $1$.